Question

n-factor of Mn(NO3)2 when it converted into MnO4–2 and NO is:

10
2
5
7

Answer 1

Answer:

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Answer 2

The converted into $(M_{n}O_{4})_{2}$ and $NO$ is 10.

Explanation:

$(M_{n}O_{4})_{2}$→$2M_{n}O_{4}^{-}$↓$2M_{n}^{2+}$

$M_{n}O_{4}^{-}$ gets reduced to $M_{n}^{2+}$ as it always does in acidic medium.

Compound has 2 atoms of $M_{n}$ in it so, factor is given by,

=change in oxidation number per atom*number of atoms per molecule or formula unit of the given reagent

$=5*2$

$=10$.